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“The height h in feet of a model rocket above the ground t seconds after lift-off is given by h(t) =𕒹t2+ 100t, for 0 ≤ t ≤ 20. When does the rocket reach its maximum height above the ground? What is its maximum height?

Differentiate the equation and put it equal to zero to find the time t for the maximum height. To find the maximum height, put that value of t into the original equation.

22/03/2018 21:35:41 | comment by Binny

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Hello, I can answer this question but would need to clarify something. There appears to be a formatting problem in the question. There is a symbol before the "t2" which appears an an empty square in my browser. Is it supposed to read "h(t) = -t^2 + 100t"? (t^2 means t squared - the ^ is used to indicate a superscri pt in typed mathematics).

27/03/2018 12:00:35 | comment by Chris

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the symbol before the `t` appears as a box, what is it meant to be showing? and does `t2` mean t squared? I think the formatting on my Chrome is incorrect

14/10/2018 19:15:38 | comment by Peter

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28 Answers

Cannot answer as this is what appears on my screen:

The height h in feet of a model rocket above the ground t seconds after lift-off is given by h(t) =𕒹;;t2+ 100t, for 0 ≤;; t ≤;; 20. When does the rocket reach its maximum height above the ground? What is its maximum height?

Can you please make sure I can see the whole equation without any unreadable symbol? I suppose it is t squared, please confirm too.

(continued) So, we start by finding the value of t at the turning point. To do this, we take the derivative of h with respect to t i.e. t = -t^2 + 100t dh/dt= -2t + 100 By setting the derivative to 0, we find the value of t at the turning point. So, if dh/dt = 0 = -2t + 100, then 2t = 100, so t = 50. We note that 50 > 20 and thus out of our range. We also note that because of our deduction earlier as our extreme value of t=20 is closest to the value of t at the turning point, this must be where our maximum occurs.

Substitute t=20 into our original equation for h: h= -20^2 + 100*20 = 1600 As a final check, use the other extreme value in our range (t=0) to see if we made a mistake. h=0 when t=0. Thus our rocket reaches its max. height when t=20 and h=1600. Suppose instead that h = -5^2 + 100t: -similar graph -dh/dt= -10t + 100 implies t=10 which is in range of values -sub. t=10 into h to find our max. height.If h(t) = 5t^2 - 100t why would the max value of h not occur at t=10?

This equation is not clearly written out; it looks like it is trying to be a simple differential equation, however h(t) doesnt really make sense. Are we multiplying the height by the time? In which case the equation will need to be rearranged to get height in terms of t. Moreover is t being squared? Again the question isnt clear. And what is the coefficient in front of t? So at present question is impossible to answer

you want to find the maximum height, that is when h(t) is the biggest value. the value of h increases as t increases. as t can only be a maximum of 20 seconds, to find the max height sub t=20 into the equation to get the value for h.

Simply differentiate the equation and let it equal to 0.From here you can find the time at max height (when the gradient of the curve = 0). Input this identified t into the original equation to find the max height.

I can`t answer the equation given to me as I believe the coefficient before the t^2 is missing

The number in front of the t^2 is missing, therefore I am unable to give you a numerical answer. However, the method that you should use is to devide the range of t by 2 and it will give you 10 since (20-0)/2=10. Now you just need to insert 10 instead of t in your equation which will give you the maximum height in feet (if the question is given in feet). I hope that answered your question.

Summary, when finding the MAX/MIN of the dependent variable (h) in a QUADRATIC EQUATION within a RANGE on the independent variable (t), we know it occurs at either an extreme value of the independent variable (t) or at a turning point, which we find by taking the derivative and setting this equal to 0. This is possible to do without sketching a graph. However, it is a good habit to get into and will be invaluable when confronted with cubic equations.

Ignoring wind resistance and assuming the rocket returns to the ground at t=20 it will follow the path of a parabola which is symmetrical. The maximum height will therefore be when t=10. Substitute t=10 into the function to get the height. However, the function in the question cannot be correct as the number before the t^2 must be negative.

To find the maxima (or minima) you need o find where the gradient of h(t) is zero. To do this, you need to differentiate h(t), see below:

dh / dt = -10t + 100 = 100 - 10t = 10(10 - t)

There is a turning point (a maxima or minima) when the differential is zero (i.e. when the gradient is zero). i.e. when dh/dt = 0

t = 10 seconds

I`m going to assume this is a maxima (if we needed to prove it was a maxima rather than a minima, we`d need to differentiate again, to see if it gave us a positive or negative result, but that is not part of the question)

I cannot see all of the equation, so, for now, I will assume that the equation given is:

h(t)= -t^2 + 100t, 0 $leq$ t $leq$ 20

If the coefficients are different the principles are still the same, so don`t sweat.

A good place to start when presented with a quadratic equation is to sketch its corresponding graph. To sketch the graph:

-solve the equation for h=0 to see where the graph crosses the x-axis

-set t=0 to see where the graph crosses the y-axis

-check: is the coefficient of the t^2 term positive or negative?

-if positive our graph shape is a smiley face

-if negative - sad face (which is what we are working with here)

We observe our graph. We notice that the highest point of the graph is at the turning point and that the further we move from this point along the x-axis the smaller the value our graph has against the y-axis. Hence, either the value of t we are looking for is at the turning point or it is the closest value within our range to the turning point. (You may wish to re-read the last two sentences and look at your graph to convince yourself that this is true). (continued)

(Sorry for lack of formatting there appears to be no way to enter a return character, put a return at each "CR")CRCR

The question hasn`t shown all characters so I guess the full equation was;CRCR

h(t) = -0.5at^2+ 100t, for 0 < t < 20. CRCR

That is a standard equation from physics. a is acceleration from gravity. The minus sign recognises that the initial velocity is upwards and gravity pulls it down.CRCR

Early in the flight (low values of t) 100t dominates and the rocket ascends. Late in the flight (high values of t) -0.5at^2 dominates and the rocket descends.CRCR

The "correct" solution is to differentiate height to get velocity;CRCR

V(t) = d h(t)/dt = -at + 100CRCR

At the highest point the rocket changes direction so v(t)=0. Assuming that a=10 m/s^2 (gravity on Earth);CRCR

0 = 100 - at CR t = 100 / 10 = 10 s CRCR

maximum height is; CRCR

h(t) = 100t - 0.5at^2 CR = 1000 - 0.5 x 10 x 100 CR = 500m CRCR

The "cheat" way to do it is to notice that the equation is specified for the time 0 < t < 20. If the rocket flight is symmetrical (true for unpowered flight in a vacuum) it will ascend for the same time that it descends and the maximum height is at the mid point t=10s. The latter is not the textbook solution but allows a quick check on your answer.

The rocket reaches its maximum height when it`s at the top of its curve, when it is momentarily still before it comes back down. when it is momentarily still, the gradient of its curve is 0, i.e. when d(h)/d(t)= 0

The formatting of the equation in you question appears corrupted, but is an equation of motion of the form d = u.t + 1/2.a.t^2 (where d is the distance, and in this case height, u is the initial speed, t is the time elapsed, and a is the acceleration which in this case is due to gravity, i.e. 9.8m/s/s). I rather think the corrected form would be d = 100 x t + 1/2 x 9.8 x t^2.

Although the equation comes from physics, the real point of the question is about turning points for a curve, which is mathematics.

The simplest, but more time consuming solution, is to tabulate values and draw a graph. However, if you have done some calculus you should be able to differentiate the equation to determine the gradient of the curve at any time t. As the maximum height will be where the differential of height with respect to time is zero you can use this to solve for when this occurs. (Note: You will need to be careful of the sign convention you use, i.e. consider which direct the distance and acceleration increases in a apply a sign accordingly.)

This is a poorly framed question. It gives no information about what happens after 20 seconds. It is curious that the unit of distance in the question is feet: in UK Maths, for many decades, distance has been measured in metres.

The narrow answer is: within the 20-second window, the rocket reaches maximum height when t = 20 and the height at that time is 2400 feet (20^2 + 100 x 20).

However, a broader answer must consider what happens beyond the 20-second window. At t = 20, the rocket will be rising at 140 feet per second. Assuming that there is no upward acceleration after that time, and assuming deceleration due to gravity of 32 feet per second per second (and assuming that air resistance is negligible), the velocity will become zero at t = 24.375; at that time, the maximum height of the rocket`s trajectory will be reached, which is 3626.25 feet.

If you consider the graph of the function which defines the height of the rocket, it had roots at both -100 and 0. We can however eliminate any aspects of the graph which are outside the range of x is less than or equal to 0 and x is more than or equal to 20. this means that the the maximum point on the graph (of which represents the maximum height of the rocket) can be clearly seen on the graph at t = 20. the maximum height of this point is h(20) = 2400.

h=102 feet, at either 20 seconds, if this is the point at which the rocket reaches its highest point. If however, 20 seconds is the time at which the rocket has returned back to the ground, we can presume that the rocket reaches its peak at 20/2=10 seconds.